Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $y = \dfrac{-6k - 6}{k + 5} \times \dfrac{k + 5}{k^2 + 9k + 8} $
Solution: First factor the quadratic. $y = \dfrac{-6k - 6}{k + 5} \times \dfrac{k + 5}{(k + 1)(k + 8)} $ Then factor out any other terms. $y = \dfrac{-6(k + 1)}{k + 5} \times \dfrac{k + 5}{(k + 1)(k + 8)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ -6(k + 1) \times (k + 5) } { (k + 5) \times (k + 1)(k + 8) } $ $y = \dfrac{ -6(k + 1)(k + 5)}{ (k + 5)(k + 1)(k + 8)} $ Notice that $(k + 5)$ and $(k + 1)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -6\cancel{(k + 1)}(k + 5)}{ (k + 5)\cancel{(k + 1)}(k + 8)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $y = \dfrac{ -6\cancel{(k + 1)}\cancel{(k + 5)}}{ \cancel{(k + 5)}\cancel{(k + 1)}(k + 8)} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $y = \dfrac{-6}{k + 8} ; \space k \neq -1 ; \space k \neq -5 $